4 sin^4 x + 3sin^2 x - 1 = 0 After I borrow the 4 I just don't know what to do afterwords..

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4 sin^4 x + 3sin^2 x - 1 = 0 After I borrow the 4 I just don't know what to do afterwords..

Mathematics
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let sin^2x = u, substitute and solve as a quadratic? i.e 4u^2 + 3u - 1 = (4u - 1)(u + 1 ) so u = 1/4 or -1, therefore sin^2(x) = 1/4 or -1, use calculator
sin^-1(SQRT(1/4) = sin^-1(1/2)
Can the quadratic formula be used if its to the 4th degree because I got the same thing, however I used factoring with borrow and payback.

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Yes... you can use the QF but I think factoring, as shown above is faster...
Also, \(\large sin^2x=\frac{1}{4} \) gives \(\large sinx=\pm \frac{1}{2} \) so you'll need to solve sinx=1/2, and also sinx=-1/2
Oh, thanks. Yes square rooting yields both signs, as squaring makes positive (negative time negative). The -1 can be ignored as it would be a complex number
yep...
So the angles would be 30 and 150, and 210 and 330?
The other one would come up with i so is it okay to ignore it since it is complex?

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